Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Main{
    public static void main(String[] args){
        int N;
        double res=0;
        Scanner in=new Scanner(System.in);
        N=in.nextInt();
        double[]p=new double[N+1];
        double[][] dp=new double[N+1][N+1];//前i个中有j个硬币朝上
        for(int i=1;i<=N;i++)p[i]=in.nextDouble();
        dp[0][0]=1;
        for(int i=1;i<=N;i++){
          dp[i][0]=dp[i-1][0]*(1-p[i]);
          for(int j=1;j<=i;j++){
            dp[i][j]=dp[i-1][j-1]*p[i]+dp[i-1][j]*(1-p[i]);
          }
        }
        for(int i=N/2+1;i<=N;i++){
          res+=dp[N][i];
        }
        System.out.println(res);
    }
}